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Question: A bar is subjected to equal and opposite forces as shown in the figure. PQRS is a plane making angle...

A bar is subjected to equal and opposite forces as shown in the figure. PQRS is a plane making angle θ with the cross-section of the bar. If the area of cross-section be ‘A’, then what is the tensile stress on PQRS

A

F / A

B

Fcosθ/ A

C

F cos2θ / A

D

F / Acosθ

Answer

F cos2θ / A

Explanation

Solution

As tensile stress = Normal forceArea=FNAN\frac{\text{Normal force}}{\text{Area}} = \frac{F_{N}}{A_{N}}

and here AN=(A/cos⁡θ)A_{N} = (A/\cos\theta), FN=F_{N} = Normal force = F cosθ

So, Tensile stress =Fcos⁡θA/cos⁡θ=Fcos⁡2θA= \frac{F\cos\theta}{A/\cos\theta} = \frac{F\cos^{2}\theta}{A}